Nowhere commutative semigroups = rectangular bands?

Question

I have recently read about Nowhere commutative semigroups, and there, they say, if $S$ is a semigroup, then these statements are equivalent:

• $S$ is nowhere commutative($ab=ba$ implies $a=b$).
• $S$ is a rectangular band(For all $a$ and $b$ in $S$, $aba = a$).

There, they used a proof with something like Green classes, but I have no idea, what they meant to be. Can you give me a proof, which is easier to understand? Any help appreciated!

Answer 1

Let $a, b \in S$. Since $a(aa) = (aa)a$, the elements $a$ and $aa$ commute and thus $a = aa$. Thus $S$ is a band. Now, $a(aba) = aaba = aba= abaa = (aba)a$, which shows that $aba$ and $a$ commute, whence $aba = a$. Thus $S$ is a rectangular band.

In the opposite direction, let $S$ be a rectangular band and suppose that $a$ and $b$ commute. Then $a = aba = baa = ba = bba = bab = b$.